---
tag: 0JP0W07Y9A
kind: statement
role: theorem
name: Fundamental theorem of algebra
aliases: [dalembert-gauss-theorem, dalemberts-theorem]
msc: [12D10]
uses: [09FTAET5Z4, 0EHKVRF4J6]
sources:
  - url: https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra
    locator: lead section
  - url: https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra
    locator: Complex-analytic proofs
same_as:
  wikidata: Q192760
  wikipedia: Fundamental theorem of algebra
---

## Statement

Let $n \ge 1$ be an integer and let $a_0, a_1, \dots, a_n$ be [[complex numbers|09FTAET5Z4]] with $a_n \neq 0$.
Let $p(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0$.
Then there is a complex number $z_0$ with $p(z_0) = 0$.

## Remarks

Proof with [[Liouville's theorem|0EHKVRF4J6]]: if $p$ has no zero, then $1/p$ is entire. Since $|p(z)| \to \infty$ as $|z| \to \infty$, $1/p$ is bounded, so it is constant, which is false for $n \ge 1$.
By induction on $n$, it follows that $p(z) = a_n (z - z_1) \cdots (z - z_n)$ for some $z_1, \dots, z_n \in \mathbb{C}$. So $p$ has exactly $n$ roots, counted with multiplicity.
The theorem holds for polynomials with real coefficients too, but the roots can be non-real: $z^2 + 1$ has the roots $i$ and $-i$.

## Explanations

### L1 human

Each polynomial equation of degree at least $1$ has a solution in the complex numbers.
So we never need to add new numbers to $\mathbb{C}$ to solve polynomial equations, unlike the step from $\mathbb{R}$ to $\mathbb{C}$ for $x^2 = -1$.